Given an array arr, the task is to count the total number of set bits in all numbers of that array arr.

Example:

Input: arr[] = {1, 2, 5, 7}
Output: 7
Explanation: Number of set bits in {1, 2, 5, 7} are {1, 1, 2, 3} respectively

Input: arr[] = {0, 4, 9, 8}
Output: 4

Approach: Follow the below steps to solve this problem:

1. Create a variable cnt to store the answer and initialize it with 0.
2. Traverse on each element of the array arr.
3. Now for each element, say x, run a loop while it's greater than 0.
4. Extract the last bit of x using (x&1) and then right shift x by a single bit.
5. Return cnt as the answer to this problem.

Below is the implementation of the above approach:

// C++ code for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the total number of set bits
// in an array of integers
int totalSetBits(vector<int>& arr)
{
    int cnt = 0;
    for (auto x : arr) {

        // While x is greater than 0
        while (x > 0) {

            // Adding last bit to cnt
            cnt += (x & 1);

            // Right shifting x by a single bit
            x >>= 1;
        }
    }
    return cnt;
}

// Driver Code
int main()
{
    vector<int> arr = { 1, 2, 5, 7 };
    cout << totalSetBits(arr);
}

// Java code for the above approach
import java.util.*;

class GFG{

// Function to count the total number of set bits
// in an array of integers
static int totalSetBits(int[] arr)
{
    int cnt = 0;
    for (int x : arr) {

        // While x is greater than 0
        while (x > 0) {

            // Adding last bit to cnt
            cnt += (x & 1);

            // Right shifting x by a single bit
            x >>= 1;
        }
    }
    return cnt;
}

// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 5, 7 };
    System.out.print(totalSetBits(arr));
}
}

// This code is contributed by shikhasingrajput 

# python code for the above approach

# Function to count the total number of set bits
# in an array of integers
def totalSetBits(arr):

    cnt = 0
    for x in arr:

        # While x is greater than 0
        while (x > 0):

            # Adding last bit to cnt
            cnt += (x & 1)

            # Right shifting x by a single bit
            x >>= 1
    return cnt

# Driver Code
if __name__ == "__main__":

    arr = [1, 2, 5, 7]
    print(totalSetBits(arr))

# This code is contributed by rakeshsahni

// C# code for the above approach
using System;

class GFG {

    // Function to count the total number of set bits
    // in an array of integers
    static int totalSetBits(int[] arr)
    {
        int cnt = 0;
        for (int x = 0; x < arr.Length; x++) {

            // While x is greater than 0
            while (arr[x] > 0) {

                // Adding last bit to cnt
                cnt += (arr[x] & 1);

                // Right shifting x by a single bit
                arr[x] >>= 1;
            }
        }
        return cnt;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 1, 2, 5, 7 };
        Console.WriteLine(totalSetBits(arr));
    }
}

// This code is contributed by ukasp.

<script>

// JavaScript code for the above approach

// Function to count the total number of set bits
// in an array of integers
function totalSetBits(arr) 
{
    let cnt = 0;
    for(let x of arr) 
    {
        
        // While x is greater than 0
        while (x > 0) 
        {
            
            // Adding last bit to cnt
            cnt += (x & 1);

            // Right shifting x by a single bit
            x >>= 1;
        }
    }
    return cnt;
}

// Driver Code
let arr = [ 1, 2, 5, 7 ];

document.write(totalSetBits(arr));

// This code is contributed by Potta Lokesh

</script>

7

Time Complexity: O(N)
Auxiliary Space: O(1)

Using bin:

• We use list comprehension to iterate through each element x in the array arr.
• For each element x, we convert it to a binary string using the built-in bin() function, then count the number of '1's in the binary representation using the count() method.
• We sum up the counts of set bits for all elements in the array using the sum() function and return the total count.

#include <iostream>
using namespace std;

// Function to count the total number of set bits in an
// array of integers
int totalSetBits(int arr[], int n)
{
    int totalSetBits = 0;
    for (int i = 0; i < n; ++i) {
        totalSetBits += __builtin_popcount(arr[i]);
    }
    return totalSetBits;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 5, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << totalSetBits(arr, n) << endl; // Output: 7
    return 0;
}

public class Solution {
    // Function to count the total number of set bits in an
    // array of integers
    public static int totalSetBits(int[] arr)
    {
        int totalSetBits = 0;
        for (int num : arr) {
            totalSetBits += Integer.bitCount(num);
        }
        return totalSetBits;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 5, 7 };
        System.out.println(totalSetBits(arr)); // Output: 7
    }
}

// This code is contributed by Akshita

# Function to count the total number of set bits in an array of integers
def total_set_bits(arr):
    return sum(bin(x).count('1') for x in arr)


# Driver Code
if __name__ == "__main__":
    arr = [1, 2, 5, 7]
    print(total_set_bits(arr))

function totalSetBits(arr) {
    let totalSetBits = 0;
    for (let num of arr) {
        totalSetBits += num.toString(2).split('1').length - 1;
    }
    return totalSetBits;
}

// Driver Code
const arr = [1, 2, 5, 7];
console.log(totalSetBits(arr)); // Output: 7

7

Time Complexity: O(N⋅M), where N is the number of elements in the array and M is the maximum number of bits in any element.

Space Complexity: O(1)